ethylene glycol, the substance used in automobile antifreeze, is composed of 38.7% c, 9.7% h, and 51.6% o by mass. its molar mass is 62.1 g/mol. determine both the empirical and molecular formula of ethylene glycol.

Respuesta :

Like all automotive antifreeze compounds, they are 38.7% carbon. So carbon is 38.7 9.7% hydrogen, 9.7 hydrogen and 51.6% oxygen. all right. Ah 1500 6% oxygen. yes. all right. Next, we need to empirically find molecular phone lines emitted from the ground.

Well, first of all, yes there are more. We have the atomic weight of carbon. 1201 Oxygen 16. The process divides each percentage by the fractional mass by 50%. So divided by 3 good points to find 38.7. This is 9.7 divided by 1, which is 9.7, and 51.6 divided by 16, which is 5157 divided by 16. I want to believe all right. The next step is to share each. youngest age. The minimum he is 3.225. If you're sharing now, we're sharing 3.225 times 3.225 is one 9.7 x 3.225. Approximately 3 and 3.225 divided by 3.225. So we have an empirical formula where carbon is 1, hydrogen is 3, and oxygen is 1. See below for 3. A small amount of Italian black hole was given. First find the molecular formula. The empirical mass is 2, 3, 16, which is 12 plus 1. So 15 plus 7 30 from g Berman. This is the molecular formula of the molecule. Therefore, we can find the sector of the mass of the molecular formula divided by the mass of the empirical formula. Remember that we are using this subdivision. Thus, for molecular  and larger LAMAs.

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